Thread: Assigning value during pointer declaration

Hi,

I am trying to understand the behavior of following code. Basically how does printf() prints the value rather than address.
Does initializing value to a pointer during declaration makes a difference when assigned from a variable?

Code:

  1 #include <stdio.h>
  2 
  3 int main() {
  4     const char *var1 = 'A';
  5     int *vint = 10;
  6     
  7     printf("\n var1 = %c", var1);
  8     printf("\n vint = %d", vint);
  9     
 10     printf("\n addr = %p", vint);
 11  
 12 }
 13

Thanks for the quick reply. If this is the case then we could do the following,

Code:

int vinit = 10

Why do we need such code? I see such code being used by many folks in many places.

Yes, you can do that. If you really want to initialise a pointer, you can then do:

Code:

 int * ptr = &vinit;

Originally Posted by pkumarn

Why do we need such code? I see such code being used by many folks in many places.

What code do you mean? Code like in your original post is definitely not commonplace, since it makes no sense on most systems.

Code you might see often is:

Code:

 char * str = "lalalala";

This will set str to the address of the lalala string. String literals are special in this regard. If you'd put your 'A' in doublequotes and left the rest of the code the same, you'd get a partial address printed from the first printf.

To dig a bit more. As you said memory is not allocated for int, then how accessing "vinit" in printf() prints 10. At first glance, vinit should be address and not value 10. Maybe i am missing something here.

Indeed -- this code has plenty of potential for disaster.

GCC will have a good moan about it:

Code:

test.c: In function ‘main’:
test.c:4:25: warning: initialization makes pointer from integer without a cast [enabled by default]
test.c:5:18: warning: initialization makes pointer from integer without a cast [enabled by default]
test.c:7:6: warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘const char *’ [-Wformat]
test.c:8:6: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat]

It might be a bit clearer to you if I add some explicit casts (I think this is one time where I do prefer C++ -- it won't let you do these assignments without explicit casts.):

Code:

     const char *var1 = (const char*)'A';
     int *vint = (int*)10;

    int value_of_vint = *vint; // !!!!

The extra line there will try to go and read from address 10.

Pointers don't automatically point to anything. If you hadn't initialised them they'd have garbage like any other local variable. To use them for anything you have to write an address to them, e.g.

Code:

  int * a = malloc(sizeof(int)); // a = addr on heap
  int i;
  int *b = &i;                 // b = addr of i, on stack
  int *c = &glob;          // address of some global 

  unsigned int *d = (unsigned int*)0x80001000;   // someplace....

You wouldn't hard code an address like that unless you knew what was there and knew the program would be able to access it (e.g. a memory mapped peripheral in an embedded system).

Thanks folks. Things are clear now.